ANDROID：删除重复字符串

Question

我在完全删除重复字符串时遇到问题List<String>String [] myno1 = new String [] {"01", "02", "03", "04", "05", "06","07", "08", "09", "10", "11", "12", "13", "14", "15"};String [] myno = new String [] {"01", "03", "15"};List<String> stringList = new ArrayList<String>(Arrays.asList(myno));List<String> stringList1 = new ArrayList<String>(Arrays.asList(myno1));stringList.addAll(stringList1);Set<String> set = new HashSet<>(stringList);stringList.clear();stringList.addAll(set);System.out.println("=== s:" +stringList);但我得到了这个：=== s:[15, 13, 14, 11, 12, 08, 09, 04, 05, 06, 24, 07, 01, 02, 03, 10]我希望结果是这样的：=== s:[13, 14, 11, 12, 08, 09, 04, 05, 06, 24, 07, 02, 10]

Answer 1

拿去：

String[] myno1 = new String[]{"01", "02", "03", "04", "05", "06", "07", 

                              "08", "09", "10", "11", "12", "13", "14", "15"};

String[] myno2 = new String[]{"01", "03", "15"};

// use LinkedHashSet to preserve order

Set<String> set1 = new LinkedHashSet<>(Arrays.asList(myno1));

Set<String> set2 = new LinkedHashSet<>(Arrays.asList(myno2));

// find duplicates

Set<String> intersection = new LinkedHashSet<>();

intersection.addAll(set1);

intersection.retainAll(set2);

// remove duplicates from both sets

Set<String> result = new LinkedHashSet<>();

result.addAll(set1);

result.addAll(set2);

result.removeAll(intersection);

System.out.println("Result: " + result);

Result: [02, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14]

Answer 2

因为您必须从另一个列表中删除项目。HashSet<>()用于从同一数组列表中删除重复项。例如，如果列表包含 15 个两次和 3 个两次，那么它将在列表中保留一次。

这是代码

foreach(String str : stringList){

stringList1.remove(str);

}

Answer 3

如果你使用 Java 8 或 +，你可以使用这个：

String[] myno1 = new String[] { "01", "02", "03", "04", "05", "06", "07", "08", "09", "10", "11", "12", "13", "14", "15" };

String[] myno = new String[] { "01", "03", "15" };

List<String> stringList = new ArrayList<>(Arrays.asList(myno));

List<String> stringList1 = new ArrayList<>(Arrays.asList(myno1));

stringList.addAll(stringList1);

List<String> newList = stringList.stream()

            .filter(string -> Collections.frequency(stringList, string) == 1)

            .collect(Collectors.toList());

System.out.println("=== s:" + newList);

不需要那么多地更改您的代码。创建了一个没有插入重复元素的新列表。输出是：

=== s:[02, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14]